There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3 , a ball is taken from the Ist bag but it shows up any other number, a ball is chosen from the II bag. Find the probability of choosing a black ball.
Since, Bag I $=\{3$ black, 4 white balls $\}$, Bag II $=\{4$ black, 3 white balls $\}$
Let $E_1$ be the event that bag $I$ is selected and $E_2$ be the event that bag $I I$ is selected.
Let $E_3$ be the event that black ball is chosen.
$\therefore \quad P\left(E_1\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$ and $P\left(E_2\right)=1-\frac{1}{3}=\frac{2}{3}$
$$\begin{array}{l} &\text { and } & P\left(E_3 / E_1\right) =\frac{3}{7} \text { and } P\left(E_3 / E_2\right)=\frac{4}{7} \\ & \therefore & P\left(E_3\right) =P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right) \\ & & =\frac{1}{3} \cdot \frac{3}{7}+\frac{2}{3} \cdot \frac{4}{7}=\frac{11}{21} \end{array}$$
There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.
$$\begin{array}{ll} \text { Let } & U_1=\{2 \text { white, } 3 \text { black balls }\} \\ & U_2=\{3 \text { white, } 2 \text { black balls }\} \\ \text { and } & U_3=\{4 \text { white, } 1 \text { black balls }\} \\ \therefore & P\left(U_1\right)=P\left(U_2\right)=P\left(U_3\right)=\frac{1}{3} \end{array}$$
Let $E_1$ be the event that a ball is chosen from urn $U_1, E_2$ be the event that a ball is chosen from urn $U_2$ and $E_3$ be the event that a ball is chosen from urn $U_3$.
Also, $$P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=1 / 3$$
Now, let $E$ be the event that white ball is drawn.
$$\therefore \quad P\left(E / E_1\right)=\frac{2}{5}, P\left(E / E_2\right)=\frac{3}{5}, P\left(E / E_3\right)=\frac{4}{5}$$
$$\begin{aligned} &\text { Now, }\\ &\begin{aligned} P\left(E_2 / E\right) & =\frac{P\left(E_2\right) \cdot P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right)} \\ & =\frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{1}{3} \cdot \frac{2}{5}+\frac{1}{3} \cdot \frac{3}{5}+\frac{1}{3} \cdot \frac{4}{5}} \\ & =\frac{\frac{3}{15}}{\frac{2}{15}+\frac{3}{15}+\frac{4}{15}}=\frac{3}{9}=\frac{1}{3} \end{aligned} \end{aligned}$$
By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99 . The probability of an healthy person diagnosed to have TB is 0.001 . In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
$$\begin{aligned} & \text { Let } E_1=\text { Event that person has TB } \\ & E_2=\text { Event that person does not have TB } \\ & E=\text { Event that the person is diagnosed to have TB } \\ & \therefore \quad P\left(E_1\right)=\frac{1}{1000}=0.001, P\left(E_2\right)=\frac{999}{1000}=0.999 \\ & \text { and } \quad P\left(E / E_1\right)=0.99 \text { and } P\left(E / E_2\right)=0.001 \\ & \therefore \quad P\left(E_1 / E\right)=\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)} \\ & =\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.001} \\ & =\frac{0.000990}{0.000990+0.000999} \\ & =\frac{990}{1989}=\frac{110}{221} \end{aligned}$$
An item is manufactured by three machines $A, B$ and $C$. Out of the total number of items manufactured during a specified period, $50 \%$ are manufactured on $A, 30 \%$ on $B$ and $20 \%$ on $C$. $2 \%$ of the items produced on $A$ and $2 \%$ of items produced on $B$ are defective and $3 \%$ of these produced on $C$ are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine $A$ ?
Let $E_1=$ Event that item is manufactured on $A$, $E_2=$ Event that an item is manufactured on $B$,
$E_3=$ Event that an item is manufactured on $C$,
Let $E$ be the event that an item is defective.
$$\therefore \quad P\left(E_1\right)=\frac{50}{100}=\frac{1}{2}, P\left(E_2\right)=\frac{30}{100}=\frac{3}{10} \text { and } P\left(E_3\right)=\frac{20}{100}=\frac{1}{5}$$
$P\left(\frac{E}{E_1}\right)=\frac{2}{100}=\frac{1}{50}, P\left(\frac{E}{E_2}\right)=\frac{2}{100}=\frac{1}{50}$ and $P\left(\frac{E}{E_3}\right)=\frac{3}{100}$
$$\begin{aligned} \therefore \quad P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{50}}{\frac{1}{2} \cdot \frac{1}{50}+\frac{3}{10} \cdot \frac{1}{50}+\frac{1}{5} \cdot \frac{3}{100}} \\ & =\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{500}+\frac{3}{500}}=\frac{\frac{1}{100}}{\frac{5+3+3}{500}}=\frac{5}{11} \end{aligned}$$
Let $X$ be a discrete random variable whose probability distribution is defined as follows.
$$P(X=x)= \begin{cases}k(x+1), & \text { for } x=1,2,3,4 \\ 2 k x, & \text { for } x=5,6,7 \\ 0, & \text { otherwise }\end{cases}$$
where, $k$ is a constant. Calculate
(i) the value of $k$.
(ii) $E(X)$.
(iii) standard deviation of $X$.
$$\begin{aligned} &P(X=x)= \begin{cases}k(x+1), & \text { for } x=1,2,3,4 \\ 2 k x, & \text { for } x=5,6,7 \\ 0, & \text { otherwise }\end{cases}\\ &\text { Thus, we have following table } \end{aligned}$$
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Otherwise |
|---|---|---|---|---|---|---|---|---|
| $P(X)$ | $2k$ | $3k$ | $4k$ | $5k$ | $10k$ | $12k$ | $14k$ | 0 |
| $XP(X)$ | $2 k$ | $6 k$ | $12 k$ | $20 k$ | $50 k$ | $72 k$ | $98k$ | 0 |
| $X^2P(X)$ | $2 k$ | $12 k$ | $36 k$ | $80k$ | $250 k$ | $432 k$ | $686k$ | 0 |
(i) Since, $\Sigma P_j=1$
$$\Rightarrow \quad k(2+3+4+5+10+12+14)=1 \Rightarrow k=\frac{1}{50}$$
$$\begin{aligned} \text{(ii)}\quad \because \quad E(X) & =\Sigma X P(X) \\ \therefore \quad E(X) & =2 k+6 k+12 k+20 k+50 k+72 k+98 k+0=260 k \\ & =260 \times \frac{1}{50}=\frac{26}{5}=5.2\quad \left[\because k=\frac{1}{50}\right] \quad\text{.... (i)} \end{aligned}$$
(iii) We know that,
$$\begin{array}{rlr} \operatorname{Var}(X) & =\left[E\left(X^2\right)\right]-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma\{X P(X)\}]^2 & \\ & =[2 k+12 k+36 k+80 k+250 k+432 k+686 k+0]-[5.2]^2 & \text { [using Eq. (i) }] \\ & =[1498 k]-27.04=\left[1498 \times \frac{1}{50}\right]-27.04 & {\left[\because k=\frac{1}{50}\right]} \\ & =29.96-27.04=2.92 & \end{array}$$
We know that, standard deviation of $X=\sqrt{\operatorname{Var}(X)}=\sqrt{2.92}=1.7088=1.7$ (approx)